part a

it can't be a time invariant because output is not same.

$ X_k[n] = \delta[n-k] $

$ \delta[n-k]\rightarrow time delay\rightarrow\delta[n-k-n_0]\rightarrow system \rightarrow (k+n_0+1)^2\delta[n-(k+n_0+1)] $

$ \delta[n-k]\rightarrow system \rightarrow Y_k[n] \rightarrow time delay \rightarrow (k+1)^2\delta[n-(k+n_0+1)] $


part b

since it is linear input would be X[n] = u[n]

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Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010